In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, -k)
(ii) (8, 1), (k, -4), (2, -5)

We know that for collinear points area of triangle = 0 ,i.e.,
\( 0 \ = \frac{1}{2}[x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \)

(i) \( 7(1-k) + 5(k+2) + 3(-2-1) \ = \ 0 \)
=> \( 7 - 7k + 5k + 10 - 9 \ = \ 0 \)
=> \( 2k \ = \ 8 \)
=> \( k \ = \ 4 \)

(ii) \( 8(-4+5) + k(-5-1) + 2(1+4) \ = \ 0 \)
=> \( 8 - 6k + 10 \ = \ 0 \)
=> \( 6k \ = \ 18 \)
=> \( k \ = \ 3 \)